Question

Let $a_1,a_2,a_3\dots \dots \dots ..$. be an arithmetic progression and $b_1,b_2,b_3,\dots \dots$ be a geometric progression. The sequence $c_1,c_2,c_3,\dots \dots \dots .$. is such that $c_n=a_n+b_n\forall n\in N$. Suppose $c_1=1,c_2=4$, $c_3=15$ and $c_4=2$.
The common ratio of geometric progression is equal to

• A. - 2
• B. - 3
• C. 2
• D. 3

Answer 1

Answer: (B)

We have
$c_1=a+b=1$ ....(1)
$c_2=(a+d)+br=4$....(2)
$c_3=(a+2d)+b{r}^2=15$....(3)
$c_4=(a+3d)+b{r}^3=2$....(4)
Now,
(2) $-(1)\Rightarrow d+b(r-1)=3$....(5)
(3) $-(2)\Rightarrow d+b\left(r^2-r\right)=11$....(6)
(4) $-(3)\Rightarrow d+b\left(r^3-r^2\right)=-13$....(7)
Also,
$(6)-(5)\Rightarrow b\left(r^2-2r+1\right)=8$....(8)
$(7)-(6)\Rightarrow b\left(r^3-2r^2+r\right)=-24$....(9)
So, $\frac{(8)}{(9)}\Rightarrow \frac{(r-1{)}^2}{r(r-1{)}^2}=\frac{1}{r}=\frac{-1}{3}\Rightarrow r=-3($ As $r\ne 1)$
$\Rightarrow a=\frac{1}{2},b=\frac{1}{2}$ and $d=5$
Clearly common ratio of geometric progresssion is $r=-3$